Question

Above figure shows a graph of $v_x$ versus $t$ for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval $t=0$ to $t=6.00s$. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant. (c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs.

Answer 1

(a) In the interval $t_i=0s$ and $t_f=6.00s$, the motorcyclist’s velocity changes from $v_i=0$ to $v_f=8.00m/s$. Then

$a=\frac{\Delta V}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}=\frac{8.0m/s-0}{6.0-0}=1.3m/s^2$

(b) Maximum positive acceleration occurs when the slope of the velocity-time curve is greatest, at $t=3s$, and is equal to the slope of the graph, approximately $(6m/s–2m/s)/(4s-2s)=2m/s^2$.

(c) The acceleration $a=0$ when the slope of the velocity-time graph is zero, which occurs at $t=6s$, and also for $t>10s$.

(d) Maximum negative acceleration occurs when the velocity-time graph has its maximum negative slope, at $t=8s$, and is equal to the slope of the graph, approximately $–1.5m/s^2$.